Problem: Find the distance between the points (10, -8) and (3, 6). {1} {2} {3} {4} {5} {6} {7} {8} {9} {10} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {\llap{-}10} {1} {2} {3} {4} {5} {6} {7} {8} {9} {10} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {\llap{-}10} (10, -8) (3, 6) $7$ $14$
Solution: Change in $x$ 10 Change in $y$ (-8) 14 The distance is the length of the hypotenuse of this right triangle. By the Pythagorean Theorem, that length is equal to: $\sqrt{7^2 + 14^2}$ $= \sqrt{245}$ $= 7\sqrt{5}$